Answer to Question #177365 in Calculus for Moel Tariburu

Question

Evaluate the integration by substitution ∫▒sin2θ/√(〖cos〗^2+16) dθ

Answer 1

We have to find value of the integral

I = $\int \dfrac{sin2 \theta}{\sqrt{cos^2{\theta}+16}}d\theta$

Using the substitution,

$sin2\theta = 2cos\theta sin\theta$

= $\int- \dfrac{-1}{\sqrt{cos^2{\theta}+16}}2sin\theta cos\theta d\theta$

Substitute $u = cos^2{\theta}+16$

$\dfrac{du}{d\theta} = -2cos\theta sin\theta$

Hence,

$d\theta = -\dfrac{du}{2cos\theta sin\theta}$

Hence integral will become

$-\int \dfrac{du}{\sqrt{u}}$

$= -2\sqrt{u}$

Using substitution $u = cos^2{\theta}+16$

We get , $I = -2\sqrt{cos^2{\theta}+16} + C$